Date: 11/21/2001 at 12:10:17
From: Doctor Pete
Subject: Re: Taylor Expansion
Hi,
This series can be found by using the generalized Binomial Theorem,
which says
(1+x)^n = a[0] + a[1]x + a[2]x^2 + ... + a[k]x^k + ...
where
a[k] = Binomial[n,k] = n!/(k!(n-k)!).
However, n is not an integer. Since the derivative of the arcsine is
(1-x^2)^(-1/2),
we see that n = -1/2, and we substitute -x^2 for x in the formula.
Then integrating term by term gives the final result. You may wonder
how to define the binomial coefficient when n is not an integer.
Since
n!/(k!(n-k)!) = n(n-1)(n-2)...(n-k+1)/k!,
it is natural to use the right-hand side when n is not an integer.
For n = -1/2, we have
Binomial[1/2,k] = (-1/2)(-3/2)(-5/2)...(-(2k-1)/2)/k!
= (-1)^k (1*3*5*...*(2k-1))/(k!2^k).
We can express this as factorials of integers by noting
(2k)! = (1*3*5*...*(2k-1))(2*4*6*...*(2k))
= (1*3*5*...*(2k-1))(2^k)(1*2*3*...*k)
= (1*3*5*...*(2k-1))(2^k)k!,
therefore
a[k] = (-1)^k (2k)!/((2^k)k!)/(k!2^k)
= (-1)^k (2k)!/(k!2^k)^2.
It follows that the k(th) term of the expansion of (1-x^2)^(-1/2) is
a[k](-x^2)^k = (2k)!/(k!2^k)^2 x^(2k).
Integrating with respect to x, we obtain the k(th) term of the
expansion of arcsin(x),
(2k)!/(k!2^k)^2 x^(2k+1)/(2k+1),
where k = 0 to infinity. It is not difficult to see that this agrees
with the orignal statement of the problem. By the uniqueness of the
series expansion, we can be assured that this is the series
representation of arcsin(x).
It is interesting to note that we did not rely on Taylor's formula,
which says that for F[x] = arcsin(x),
a[k] = F(k)[0]/k!,
where F(k)[x] is the k(th) derivative of F, evaluated at x.
- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/